Question: $h(x) = 5x^{2}+3x+f(x)$ $g(t) = 2t^{3}-6t^{2}+5t-2-5(f(t))$ $f(x) = -6x^{2}+2x$ $ g(h(5)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(5)$ . Then we'll know what to plug into the outer function. $h(5) = 5(5^{2})+(3)(5)+f(5)$ To solve for the value of $h$ , we need to solve for the value of $f(5)$ $f(5) = -6(5^{2})+(2)(5)$ $f(5) = -140$ That means $h(5) = 5(5^{2})+(3)(5)-140$ $h(5) = 0$ Now we know that $h(5) = 0$ . Let's solve for $g(h(5))$ , which is $g(0)$ $g(0) = 2(0^{3})-6(0^{2})+(5)(0)-2-5(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = -6(0^{2})+(2)(0)$ $f(0) = 0$ That means $g(0) = 2(0^{3})-6(0^{2})+(5)(0)-2+(-5)(0)$ $g(0) = -2$